3.598 \(\int x^5 (a+b x^2)^2 \sqrt{c+d x^2} \, dx\)

Optimal. Leaf size=157 \[ \frac{\left (c+d x^2\right )^{7/2} \left (a^2 d^2-6 a b c d+6 b^2 c^2\right )}{7 d^5}+\frac{c^2 \left (c+d x^2\right )^{3/2} (b c-a d)^2}{3 d^5}-\frac{2 b \left (c+d x^2\right )^{9/2} (2 b c-a d)}{9 d^5}-\frac{2 c \left (c+d x^2\right )^{5/2} (b c-a d) (2 b c-a d)}{5 d^5}+\frac{b^2 \left (c+d x^2\right )^{11/2}}{11 d^5} \]

[Out]

(c^2*(b*c - a*d)^2*(c + d*x^2)^(3/2))/(3*d^5) - (2*c*(b*c - a*d)*(2*b*c - a*d)*(c + d*x^2)^(5/2))/(5*d^5) + ((
6*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*(c + d*x^2)^(7/2))/(7*d^5) - (2*b*(2*b*c - a*d)*(c + d*x^2)^(9/2))/(9*d^5) +
(b^2*(c + d*x^2)^(11/2))/(11*d^5)

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Rubi [A]  time = 0.129136, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {446, 88} \[ \frac{\left (c+d x^2\right )^{7/2} \left (a^2 d^2-6 a b c d+6 b^2 c^2\right )}{7 d^5}+\frac{c^2 \left (c+d x^2\right )^{3/2} (b c-a d)^2}{3 d^5}-\frac{2 b \left (c+d x^2\right )^{9/2} (2 b c-a d)}{9 d^5}-\frac{2 c \left (c+d x^2\right )^{5/2} (b c-a d) (2 b c-a d)}{5 d^5}+\frac{b^2 \left (c+d x^2\right )^{11/2}}{11 d^5} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^2)^2*Sqrt[c + d*x^2],x]

[Out]

(c^2*(b*c - a*d)^2*(c + d*x^2)^(3/2))/(3*d^5) - (2*c*(b*c - a*d)*(2*b*c - a*d)*(c + d*x^2)^(5/2))/(5*d^5) + ((
6*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*(c + d*x^2)^(7/2))/(7*d^5) - (2*b*(2*b*c - a*d)*(c + d*x^2)^(9/2))/(9*d^5) +
(b^2*(c + d*x^2)^(11/2))/(11*d^5)

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int x^5 \left (a+b x^2\right )^2 \sqrt{c+d x^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 (a+b x)^2 \sqrt{c+d x} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{c^2 (b c-a d)^2 \sqrt{c+d x}}{d^4}+\frac{2 c (b c-a d) (-2 b c+a d) (c+d x)^{3/2}}{d^4}+\frac{\left (6 b^2 c^2-6 a b c d+a^2 d^2\right ) (c+d x)^{5/2}}{d^4}-\frac{2 b (2 b c-a d) (c+d x)^{7/2}}{d^4}+\frac{b^2 (c+d x)^{9/2}}{d^4}\right ) \, dx,x,x^2\right )\\ &=\frac{c^2 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}{3 d^5}-\frac{2 c (b c-a d) (2 b c-a d) \left (c+d x^2\right )^{5/2}}{5 d^5}+\frac{\left (6 b^2 c^2-6 a b c d+a^2 d^2\right ) \left (c+d x^2\right )^{7/2}}{7 d^5}-\frac{2 b (2 b c-a d) \left (c+d x^2\right )^{9/2}}{9 d^5}+\frac{b^2 \left (c+d x^2\right )^{11/2}}{11 d^5}\\ \end{align*}

Mathematica [A]  time = 0.0921675, size = 132, normalized size = 0.84 \[ \frac{\left (c+d x^2\right )^{3/2} \left (33 a^2 d^2 \left (8 c^2-12 c d x^2+15 d^2 x^4\right )+22 a b d \left (24 c^2 d x^2-16 c^3-30 c d^2 x^4+35 d^3 x^6\right )+b^2 \left (240 c^2 d^2 x^4-192 c^3 d x^2+128 c^4-280 c d^3 x^6+315 d^4 x^8\right )\right )}{3465 d^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^2)^2*Sqrt[c + d*x^2],x]

[Out]

((c + d*x^2)^(3/2)*(33*a^2*d^2*(8*c^2 - 12*c*d*x^2 + 15*d^2*x^4) + 22*a*b*d*(-16*c^3 + 24*c^2*d*x^2 - 30*c*d^2
*x^4 + 35*d^3*x^6) + b^2*(128*c^4 - 192*c^3*d*x^2 + 240*c^2*d^2*x^4 - 280*c*d^3*x^6 + 315*d^4*x^8)))/(3465*d^5
)

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Maple [A]  time = 0.008, size = 149, normalized size = 1. \begin{align*}{\frac{315\,{b}^{2}{x}^{8}{d}^{4}+770\,ab{d}^{4}{x}^{6}-280\,{b}^{2}c{d}^{3}{x}^{6}+495\,{a}^{2}{d}^{4}{x}^{4}-660\,abc{d}^{3}{x}^{4}+240\,{b}^{2}{c}^{2}{d}^{2}{x}^{4}-396\,{a}^{2}c{d}^{3}{x}^{2}+528\,ab{c}^{2}{d}^{2}{x}^{2}-192\,{b}^{2}{c}^{3}d{x}^{2}+264\,{a}^{2}{c}^{2}{d}^{2}-352\,ab{c}^{3}d+128\,{b}^{2}{c}^{4}}{3465\,{d}^{5}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^2+a)^2*(d*x^2+c)^(1/2),x)

[Out]

1/3465*(d*x^2+c)^(3/2)*(315*b^2*d^4*x^8+770*a*b*d^4*x^6-280*b^2*c*d^3*x^6+495*a^2*d^4*x^4-660*a*b*c*d^3*x^4+24
0*b^2*c^2*d^2*x^4-396*a^2*c*d^3*x^2+528*a*b*c^2*d^2*x^2-192*b^2*c^3*d*x^2+264*a^2*c^2*d^2-352*a*b*c^3*d+128*b^
2*c^4)/d^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^2*(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61442, size = 398, normalized size = 2.54 \begin{align*} \frac{{\left (315 \, b^{2} d^{5} x^{10} + 35 \,{\left (b^{2} c d^{4} + 22 \, a b d^{5}\right )} x^{8} + 128 \, b^{2} c^{5} - 352 \, a b c^{4} d + 264 \, a^{2} c^{3} d^{2} - 5 \,{\left (8 \, b^{2} c^{2} d^{3} - 22 \, a b c d^{4} - 99 \, a^{2} d^{5}\right )} x^{6} + 3 \,{\left (16 \, b^{2} c^{3} d^{2} - 44 \, a b c^{2} d^{3} + 33 \, a^{2} c d^{4}\right )} x^{4} - 4 \,{\left (16 \, b^{2} c^{4} d - 44 \, a b c^{3} d^{2} + 33 \, a^{2} c^{2} d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{3465 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^2*(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

1/3465*(315*b^2*d^5*x^10 + 35*(b^2*c*d^4 + 22*a*b*d^5)*x^8 + 128*b^2*c^5 - 352*a*b*c^4*d + 264*a^2*c^3*d^2 - 5
*(8*b^2*c^2*d^3 - 22*a*b*c*d^4 - 99*a^2*d^5)*x^6 + 3*(16*b^2*c^3*d^2 - 44*a*b*c^2*d^3 + 33*a^2*c*d^4)*x^4 - 4*
(16*b^2*c^4*d - 44*a*b*c^3*d^2 + 33*a^2*c^2*d^3)*x^2)*sqrt(d*x^2 + c)/d^5

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Sympy [A]  time = 2.74027, size = 389, normalized size = 2.48 \begin{align*} \begin{cases} \frac{8 a^{2} c^{3} \sqrt{c + d x^{2}}}{105 d^{3}} - \frac{4 a^{2} c^{2} x^{2} \sqrt{c + d x^{2}}}{105 d^{2}} + \frac{a^{2} c x^{4} \sqrt{c + d x^{2}}}{35 d} + \frac{a^{2} x^{6} \sqrt{c + d x^{2}}}{7} - \frac{32 a b c^{4} \sqrt{c + d x^{2}}}{315 d^{4}} + \frac{16 a b c^{3} x^{2} \sqrt{c + d x^{2}}}{315 d^{3}} - \frac{4 a b c^{2} x^{4} \sqrt{c + d x^{2}}}{105 d^{2}} + \frac{2 a b c x^{6} \sqrt{c + d x^{2}}}{63 d} + \frac{2 a b x^{8} \sqrt{c + d x^{2}}}{9} + \frac{128 b^{2} c^{5} \sqrt{c + d x^{2}}}{3465 d^{5}} - \frac{64 b^{2} c^{4} x^{2} \sqrt{c + d x^{2}}}{3465 d^{4}} + \frac{16 b^{2} c^{3} x^{4} \sqrt{c + d x^{2}}}{1155 d^{3}} - \frac{8 b^{2} c^{2} x^{6} \sqrt{c + d x^{2}}}{693 d^{2}} + \frac{b^{2} c x^{8} \sqrt{c + d x^{2}}}{99 d} + \frac{b^{2} x^{10} \sqrt{c + d x^{2}}}{11} & \text{for}\: d \neq 0 \\\sqrt{c} \left (\frac{a^{2} x^{6}}{6} + \frac{a b x^{8}}{4} + \frac{b^{2} x^{10}}{10}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**2+a)**2*(d*x**2+c)**(1/2),x)

[Out]

Piecewise((8*a**2*c**3*sqrt(c + d*x**2)/(105*d**3) - 4*a**2*c**2*x**2*sqrt(c + d*x**2)/(105*d**2) + a**2*c*x**
4*sqrt(c + d*x**2)/(35*d) + a**2*x**6*sqrt(c + d*x**2)/7 - 32*a*b*c**4*sqrt(c + d*x**2)/(315*d**4) + 16*a*b*c*
*3*x**2*sqrt(c + d*x**2)/(315*d**3) - 4*a*b*c**2*x**4*sqrt(c + d*x**2)/(105*d**2) + 2*a*b*c*x**6*sqrt(c + d*x*
*2)/(63*d) + 2*a*b*x**8*sqrt(c + d*x**2)/9 + 128*b**2*c**5*sqrt(c + d*x**2)/(3465*d**5) - 64*b**2*c**4*x**2*sq
rt(c + d*x**2)/(3465*d**4) + 16*b**2*c**3*x**4*sqrt(c + d*x**2)/(1155*d**3) - 8*b**2*c**2*x**6*sqrt(c + d*x**2
)/(693*d**2) + b**2*c*x**8*sqrt(c + d*x**2)/(99*d) + b**2*x**10*sqrt(c + d*x**2)/11, Ne(d, 0)), (sqrt(c)*(a**2
*x**6/6 + a*b*x**8/4 + b**2*x**10/10), True))

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Giac [A]  time = 1.13152, size = 248, normalized size = 1.58 \begin{align*} \frac{\frac{33 \,{\left (15 \,{\left (d x^{2} + c\right )}^{\frac{7}{2}} - 42 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} c + 35 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} c^{2}\right )} a^{2}}{d^{2}} + \frac{22 \,{\left (35 \,{\left (d x^{2} + c\right )}^{\frac{9}{2}} - 135 \,{\left (d x^{2} + c\right )}^{\frac{7}{2}} c + 189 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} c^{2} - 105 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} c^{3}\right )} a b}{d^{3}} + \frac{{\left (315 \,{\left (d x^{2} + c\right )}^{\frac{11}{2}} - 1540 \,{\left (d x^{2} + c\right )}^{\frac{9}{2}} c + 2970 \,{\left (d x^{2} + c\right )}^{\frac{7}{2}} c^{2} - 2772 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} c^{3} + 1155 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} c^{4}\right )} b^{2}}{d^{4}}}{3465 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^2*(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/3465*(33*(15*(d*x^2 + c)^(7/2) - 42*(d*x^2 + c)^(5/2)*c + 35*(d*x^2 + c)^(3/2)*c^2)*a^2/d^2 + 22*(35*(d*x^2
+ c)^(9/2) - 135*(d*x^2 + c)^(7/2)*c + 189*(d*x^2 + c)^(5/2)*c^2 - 105*(d*x^2 + c)^(3/2)*c^3)*a*b/d^3 + (315*(
d*x^2 + c)^(11/2) - 1540*(d*x^2 + c)^(9/2)*c + 2970*(d*x^2 + c)^(7/2)*c^2 - 2772*(d*x^2 + c)^(5/2)*c^3 + 1155*
(d*x^2 + c)^(3/2)*c^4)*b^2/d^4)/d